anonymous
  • anonymous
Trig identity.. I don't understand this one at all
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
how do i make the left side into one big fraction
anonymous
  • anonymous
\[\huge \frac{ 1 }{ sinx+1 } - \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ 1-\sin^2x}\]
anonymous
  • anonymous
typed it wrong wooops

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
get common denominators on the left side: \(\huge \frac{ 1 }{ sinx+1 }\cdot (\color {red}{\frac{ sinx-1 }{ sinx-1 }}) - \frac{ 1 }{ sinx-1 }\cdot( \color {red}{\frac{ sinx+1 }{ sinx+1 }}) = \frac{ 2 }{ 1-\sin^2x} \)
anonymous
  • anonymous
so the left side in simplified form will be: \(\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1} \) can you simplify this?
anonymous
  • anonymous
no that's where i get stuck how would i subtract the numerator
anonymous
  • anonymous
just use distributive property to get rid of the parenthesis...
anonymous
  • anonymous
@dpaInc i get this \[\huge \frac{ \sin^2x+sinx-sinx-1 }{ sinx^2-1 }\]
anonymous
  • anonymous
how did you get that? all you're doing is subtracting....: \(\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1} \) =\(\huge \frac{1\cdot (sinx-1)-1\cdot (sinx+1)}{sin^2x-1} \) =\(\huge \frac{sinx-1-sinx-1}{sin^2x-1} \) =\(\huge \frac{-2}{sin^2x-1} \) =\(\huge \frac{2}{1-sin^2x} \)
anonymous
  • anonymous
sinx-1-sinx-1 = -2 is that a trig rule or what ? :S

Looking for something else?

Not the answer you are looking for? Search for more explanations.